\(A=\sqrt{9x^2-6x+1}+\sqrt{y^2-4y+5}\)
\(A=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(y-2\right)^2+1}\ge1\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=2\end{matrix}\right.\)
Ta có :
\(\sqrt{9x^2-6x+1}+\sqrt{y^2-4y+5}=1\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}+\sqrt{\left(y-2\right)^2+1}=1\)
\(\Leftrightarrow\left|3x-1\right|+\left|y-2\right|=0\)
\(\left\{{}\begin{matrix}3x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=2\end{matrix}\right.\)
Kết luận ..............
Ta có:
\(\sqrt{9x^2-6x+1}+\sqrt{y^2-4y+5}=1\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}+\sqrt{\left(y-2\right)^2+1}=1\)
Vì \(\sqrt{\left(3x-1\right)^2}\ge0\) và \(\sqrt{\left(y-2\right)^2+1}\ge1\)
nên \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(y-2\right)^2+1}\ge1\)
Vậy phương trình tương đương với \(\left\{{}\begin{matrix}\sqrt{\left(3x-1\right)^2}=0\\\sqrt{\left(y-2\right)^2+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=2\end{matrix}\right.\)
